a) 2M+ 2nH2O--->2M(OH)n +nH2
Ta có
nH2=0,482=0,24(mol)H2=0,482=0,24(mol)
Theo pthh
nM=2nnH2=0,48nmolM=2nnH2=0,48nmol
=> MM=3,33:0,48n=3,33n0,48=7M=3,33:0,48n=3,33n0,48=7
Ta có
Vậy M là Li
b)mH2O=100.1=100(g)H2O=100.1=100(g)
Theo pthh
nLiOH=nH2=0,24(mol)LiOH=nH2=0,24(mol)
mLiOH=0,24.24=5,76(g)LiOH=0,24.24=5,76(g)
C%=5,76100+3,33.0,48.100%=5,6%